And a made-up binary operation on D is defined as follows: (a+b•)(c+d•)= ac+(ad+bc)•

For example, (2+3•)(-3+5•)= (-6+1•)
You are not allowed to combine (-6+1•) into -5• because they are not like terms. you are allowed to combine like terms, however, like this:
(a+b•)+(c+d•) = a+c+(b+d•)

So the question is:
Solve the quadratic equation $x^2$-2x+12•=0

I'm very confused about the 12• and binary operation part. Should the quadratic formula be used here? How would you solve it?

What, and why, is that dot to the right of some numbers?
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DonAntonioDec 1 '13 at 20:35

the dot doesn't represent multiplication or anything, it's just a symbol to differentiate like terms. it's similar to how 2q is different from 2 so you can't combine 2q+2, but q doesn't represent any value. So in this case the • is used to create the binary operation
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gticecream8Dec 1 '13 at 20:37

1 Answer
1

Let's denote the $\bullet$ rather by $q$, even if that doesn't represent any real number as value. So that, $q:=0+1\!\bullet$.

Now we have $q^2=q\cdot q=0+0\bullet=0$, and basically that implies the whole multiplication (just the same way as $i^2=-1$ and linearity generates the multiplication for complex numbers).

We have to solve $x^2-2x+12q=0$. Write up $x$ as $x=a+bq$ then we have
$x^2=a^2+2abq$, so what is needed is:
$$a^2-2a+(2ab-2b)q=-12q$$
Looking at the 'coordinates' on both sides, we need $a^2-2a=0$ and $2b(a-1)=-12$.

am I allowed to solve those two "coordinates" to find a = 0 and plug a into the second equation to find b=6 ? (excluding all the q's because the q's are confusing to solve with and I can just add them in later.) I'm confused as to what x would end up being, since it represents (a+b). As in, what would the final answer be for x
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gticecream8Dec 1 '13 at 21:07

1

Yes, either $a=0$ or $a=2$ and the corresponding $b$ values are $6$ and $-6$, meaning two solutions: $x=6\bullet$ and $x=2-6\bullet$. (For example, for $x=6\bullet$, its square is $0$, so $x^2-2x=-2x=-12\bullet$ indeed.)
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BerciDec 1 '13 at 21:17

a similar question that relates to the system but not to the quadratic equation. Is it possible to use the created binary operation to get an answer of (5+0q) where q stands in for ∙ ? I've been trying for awhile but 5 seems prime in this system, though I know it is definitely not.
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gticecream8Dec 2 '13 at 4:57